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Well, how to calculate sum from 1 to 5000 ? Instead of looping from 1 to 5000, you compute (1 + 5000) * 5000 / 2 This is what "understand problem first" actua
by lanna 3y ago
Well, how to calculate sum from 1 to 5000 ?
Instead of looping from 1 to 5000, you compute (1 + 5000) * 5000 / 2
This is what "understand problem first" actually means
- roelschroeven 3y agoBut now you can run into problems with integer overflow. It can be solved: sum(n) = is_even(n) ? n/2 * (n+1) : (n+1)/2 * n because of one n and n+1 will always be even. But it show that unfortunately things are often not as straightforward as they first appear.